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3 years ago

15

a math test is worth 100 points and has 30 problems. Each problem is worth either 3 points or 4 points, how many 4-point problem

s ate there?

1 answer:

sveta [45]3 years ago

3 0

Answer:

13

Step-by-step explanation:

100 total points

3 4

9 8

12 12

15 16

18 20

21 24

24 28

27 32

30 36

33 40

36 44

39 48

42 52

45

48

You might be interested in

Find the missing angle.<br> Enter the number that belongs in the green box.<br> Enter

Ivenika [448]

Answer:

35 degrees

Step-by-step explanation:

The sum of all angles in a triangle equals to 180

To find angle z, just subtract 55 and 90 from 180

180-55-90= 35

Angle z = 35 degrees

Hope this helps

8 0

3 years ago

0.5a − 0.3 = 5 (solve for a)

Alla [95]

0.5a - 0.3 = 5

Add 0.3 to both sides:

0.5a = 5.3

Divide both sides by 0.5:

a = 10.6

3 0

3 years ago

Read 2 more answers

In 1982 Abby’s mother scored at the 93rd percentile in the math SAT exam. In 1982 the mean score was 503 and the variance of the

oksian1 [2.3K]

Answer:

The percentle for Abby's score was the 89.62nd percentile.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Abby's mom score:

93rd percentile in the math SAT exam. In 1982 the mean score was 503 and the variance of the scores was 9604.

93rd percentile. X when Z has a pvalue of 0.93. So X when Z = 1.476.

$\mu = 503, \sigma = \sqrt{9604} = 98$

So

$Z = \frac{X - \mu}{\sigma}$

$1.476 = \frac{X - 503}{98}$

$X - 503 = 1.476*98$

$X = 648$

Abby's score

She scored 648.

$\mu = 521 \sigma = \sqrt{10201} = 101$

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{648 - 521}{101}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

The percentle for Abby's score was the 89.62nd percentile.

3 0

3 years ago

CAN SOMEBODY PLEASE HELP ME NOW I AM IN A RUSH!!!!!!!!!!!!!!!!!!!

cupoosta [38]

Answer:

i’m pretty sure it is 3 but i could be wrong

Step-by-step explanation:

3 0

3 years ago

In a certain region, about 6% of a city's population moves to the surrounding suburbs each year, and about 4% of the suburban po

Sedbober [7]

Answer:

City @ 2017 = 8,920,800

Suburbs @ 2017 = 1, 897, 200

Step-by-step explanation:

Solution:

- Let p_c be the population in the city ( in a given year ) and p_s is the population in the suburbs ( in a given year ) . The first sentence tell us that populations p_c' and p_s' for next year would be:

0.94*p_c + 0.04*p_s = p_c'

0.06*p_c + 0.96*p_s = p_s'

- Assuming 6% moved while remaining 94% remained settled at the time of migrations.

- The matrix representation is as follows:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}p_c\\p_s\end{array}\right] = \left[\begin{array}{c}p_c'\\p_s'\end{array}\right]$

- In the sequence for where x_k denotes population of kth year and x_k+1 denotes population of x_k+1 year. We have:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1$

- Let x_o be the populations defined given as 10,000,000 and 800,000 respectively for city and suburbs. We will have a population x_1 as a vector for year 2016 as follows:

$\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1$

- To get the population in year 2017 we will multiply the migration matrix to the population vector x_1 in 2016 to obtain x_2.

$x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o$

- Where,

$x_o = \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]$

- The population in 2017 x_2 would be:

$x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]$

5 0

4 years ago